##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics

# Navigation Problems - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I want to talk about a special kind of vector problem called navigation problems. Let’s start with an easy one. A plane was travelling 480 miles per hour at a heading of 70 degrees. The wind is 75 miles an hour from 160 degrees. Find the course and ground speed.

Now I’m going to draw a diagram and explain what this term mean but first of all, remember that whenever we measure the any direction on a navigation problem, the convention is to measure clockwise from north. So it’s different from the way we know we measure angles.

So let me draw a vertical line representing north. And so my 70 degree heading is going to be something like this. And remember that heading is the direction that the plane is pointing in. So let’s say that this is our plane, it's actually trying to fly in this direction 70 degrees from north. And the magnitude of this vector is 480 miles per hour, that’s the air speed, so 480. That’s the air speed and the heading, now what about the wind? The wind is 75 miles per hour, from 160 degrees. So they usually tell you where the wind is blowing from not the direction it's blowing to and that’s really important.

From 160 degrees, we're a little short of 180, so let’s say something like this. It’s coming from that direction so I’m going to draw a little dashed line. This is the wind vector though. Now this is the direction the wind is coming from, 160 degrees, and the magnitude of the wind is 75. So I'll put the 75 here.

Now you can see that as the plane is flying in this direction, the wind is trying to blow it off course. And we want to find is its true course, and ground speed. So the wind will have some effect on the actual ground speed as well. Remember that this speed is the speed of the airplane with respect to the air not respect to the ground. Now remember when you are adding vectors, you can move any vector anywhere else you want. So I’m going to move this vector over here, so that I can add them. I need to move them head to tail. It doesn’t have to be a perfect drawing I’m just going to copy the wind vector.

And then I want to represent the sum of these two, the resultant, and that’s going to look something like this. And this sum represents both the ground speed and the course. The ground speed is the magnitude of this vector and the course is going to be the angle of this vector makes with north. So I’m going to find this in the end, this will be the course. Now first of all, in order to solve this triangle, we do have a triangle here we have two sides. I need at least one more piece of information. Now what can I get?

Well I can get an angle I’ll show you how. Notice that this angle is 70 degrees and this angle in blue is 160 degrees, that means the difference which is this angle, is 90 degrees, a right angle. Now also notice that this line is parallel to this line. So I have two parallel lines cut by a transversal that means all frontier angles are going to be congruent. That means this is going to be 90 degrees. So now I know everything I need to.

I’ve got a right triangle with one leg 480, one leg 75. So right away I can find the magnitude of the ground speed. So I’ll call that G. G is going to be the magnitude of the ground speed. So observe that 480² plus 75² equals G². So I’m going to use my calculator, 480² plus 75² and that’s 236025 that’s G squared. So G is the square root of that and its 485.8 approximately. And this will be in the same units, miles per hour.

Now remember the original air speed was 480, so it gets a little bump from the wind. Now what’s the course? What I want to do is I’m going to solve for this angle. I can solve for this angle using right angle trigonometry, and then I can subtract that angle from the 70 degree heading angle and that will give me the course.

So first of all, let’s observe that in this right triangle, the tangent of theta, is side opposite over side adjacent, so it's 75 over 480. Tan-theta equals 75 over 480. So theta is inverse tangent of 75 over 480. Again, I'll use my calculator to approximate this, inverse tangent 75 over 480. 8.9 degrees approximately. And that means that the course is 70 degrees minus that. So 70 minus, answer, the course is approximately 61.1 degrees.

So in the end, our course is 61.1 degrees from north, and our ground speed is 485.8 miles per hour.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete